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  1. MMK95

    48 HOURS LEFT FOR 0606 ADD MATHS.....HOW IS UR PREPARATION?????

    Alhumdulillah the preparations are good....................... In-Sha-Allah we will get an easy paper tomorrow............... :)
  2. MMK95

    Additional Mathematics help!

    Sorry my bad..... :) here we have to find the distance traveled in the 12th second not 12 seconds, so subtract t=11 form t=12 and you have the answer.... :)
  3. MMK95

    ADDITIONAL MATHEMATICS

    My school told me that if I convince 6 students to take add maths then they would arrange the classes but unfortunately I could not and we share the same fate...... :( ............... as for that topic I would say that dont get stressed over it... I have observed that the question of this topic...
  4. MMK95

    Additional Mathematics help!

    For the first part, just put t = 12 in the equation and find s ( displacement = distance).... as for the second part derivate the equation s because the derivative of displacement give the equation of velocity.... at instantaneous rest velocity is always equal to 0, so the equation formed would...
  5. MMK95

    ADDITIONAL MATHEMATICS

    Check this one.... I hope you will find this helpful..... http://drtayeb.files.wordpress.com/2011/05/relatvie-velocity.pdf
  6. MMK95

    ADDITIONAL MATHEMATICS

    I know that the papers are same and I have solved many of them, but the questions in o levels papers are harder if compared to igcse.... :)
  7. MMK95

    ADDITIONAL MATHEMATICS

    The syllabus is the same but o levels papers are lot harder than igcse
  8. MMK95

    ADDITIONAL MATHEMATICS

    can anyone explain Q 7 (b) on page 233 of the book by Hoo Soo Thong............ thanks in advance :)
  9. MMK95

    ADDITIONAL MATHEMATICS

    I think yes......
  10. MMK95

    Physics, Chemistry and Biology: Post your doubts here!

    Benzene is a hydrocarbon 100 - 92.3 = 7.7 92.3/12 : 7.7/1 7.1 : 7.7 1:1 Q7 1 mole of butane reacts with 6.5 moles of Oxygen 1 cm^3 of butane reacts with 6.5 cm^3 of oxygen ( Mole ratio and volume ratio are same ) 10 cm^3 reacts with 65 cm^3 Oxygen left 100 - 65 = 35 1 cm^3 of butane forms...
  11. MMK95

    Mathematics: Post your doubts here!

    mean = (sum of fx)/(sum of f) 2.95 = sum of fx/60 sum of fx = 60 * 2.95 = 177 to get the fx of first 50 rolls multiply you answer of c with 50 2.96 * 50 = 148 fx for the 10 rolls = 177 - 148 = 29 mean = sum of fx/sum of f = 29/10...
  12. MMK95

    Mathematics: Post your doubts here!

    Area of ABC = Area of ABD + Area of DBC 40 = [1/2 (x+6)(x+1)] + [1/2 (x+2)(x+1)] 40 = (x^2 +7x + 6)/2 + (x^2 +3x + 2)/2 40 = (2x^2 + 10x + 8)/2 40 = x^2 +5x + 4 x^2 +5x +4 - 40 = 0 x^2 + 5x - 36 = 0 (shown)
  13. MMK95

    Mathematics: Post your doubts here!

    Which part??????????
  14. MMK95

    Mathematics: Post your doubts here!

    use this formula for similar figures (V1/V2) = (Side1/Side2)^3 by assuming that the volume of the removed cone is x and the answer of Q10 a is 2035.8 (2035.8/x) = (24/8)^3 (2035.8/x) = (13824/512) x = 2035.8/27 x = 75.4 Volume of the remaining solid = 2035.8 - 75.4...
  15. MMK95

    Physics, Chemistry and Biology: Post your doubts here!

    exa Exactly what I needed...... Thanks :)
  16. MMK95

    Mathematics: Post your doubts here!

    if you ind any question abt it, then pls ask but I think notnek01 is right.......
  17. MMK95

    Mathematics: Post your doubts here!

    pls give the link of the question
  18. MMK95

    Mathematics: Post your doubts here!

    if the equation is y=3x+5, then the gradient of this equation is 3. The gradient of the perpendicular would be -1/3, now the midpoint of the line y=3x+5 is needed. I am sure that it is given in the question
  19. MMK95

    Mathematics: Post your doubts here!

    if the gradient of the given line is m, then the gradient of its perpendicular would be -1/m. Take the common point and use it to find c
  20. MMK95

    Physics, Chemistry and Biology: Post your doubts here!

    Density of copper is 8.9gm/cm^3 while density of aqueous zinc sulfate in 3.9gm/cm^3, so Cu would be at the bottom......... :)
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